Calculus: Early Transcendentals, Chapter 7, 7.1, Section 7.1, Problem 22

You need to use integration by parts, such that:

int udv = uv - int v du
u = (arcsin^2(x))=> du = (2arcsin x)/(sqrt(1-x^2))dx
dv = 1 => v = x
int (arcsin^2(x)) dx = x*(arcsin^2(x)) - 2 int arcsin x*x/(sqrt(1-x^2))dx
You need to solve the integral int arcsin x*(x/(sqrt(1-x^2)))dx using substitution arcsin x= t , such that:
arcsin x = t => (dx)/(sqrt(1-x^2))= dt
int arcsin x*(x/(sqrt(1-x^2)))dx = int t*sin t dt
You need to use integration by parts to solve int t*sin t dt , such that:
u = t => du = dt
dv = sin t => v = -cos t
int t*sin t dt = -t*cos t + int cos t dt
int t*sin t dt = -t*cos t + sint + c
Replacing back the variable yields:
int arcsin x*(x/(sqrt(1-x^2)))dx = -arcsin x*cos(arcsin x) + x + c
int (arcsin^2(x)) dx = x*(arcsin^2(x)) - 2( -arcsin x*cos(arcsin x) + x) + C
int (arcsin^2(x)) dx = x*(arcsin^2(x)) + 2arcsin x*cos(arcsin x) - 2x + c
Hence, evaluating the indefinite integral, using parts and substitution, yields int (arcsin^2(x)) dx = x*(arcsin^2(x)) + 2arcsin x*cos(arcsin x) - 2x + c.